Question

\[\int x^2 \tan^{- 1} x\ dx\]

Answer 1

\[\text{We have}, \]
\[I = \int x^2 \tan^{- 1} x \text{ dx }\]
\[\text{Considering} \tan^{- 1}  \text{  x   as first function and  x}^2 \text{as second function}\]
\[I = \tan^{- 1} x\frac{x^3}{3} - \int\left( \frac{1}{1 + x^2} \times \frac{x^3}{3} \right)dx\]
\[ = \tan^{- 1} x\frac{x^3}{3} - \frac{1}{3}\int\frac{x^3 dx}{1 + x^2}\]
\[ = \tan^{- 1} x\frac{x^3}{3} - \frac{1}{3}\int\left( \frac{x^2 x}{1 + x^2} \right)dx\]
\[\text{ Putting 1 + x}^2 = t\]
\[ \Rightarrow x^2 = t - 1\]
\[ \Rightarrow \text{ 2x dx = dt}\]
\[ \Rightarrow x\text{  dx }= \frac{dt}{2}\]
\[ \therefore I = \tan^{- 1} x\frac{x^3}{3} - \frac{1}{6}\int\left( \frac{t - 1}{t} \right)dt\]
\[ = \frac{x^3}{3} \tan^{- 1} x - \frac{1}{6}\int dt + \frac{1}{6}\int\frac{dt}{t}\]
\[ = \frac{x^3}{3} \tan^{- 1} x - \frac{1}{6}t + \frac{1}{6}\text{ log }\left| t \right| + C\]
\[ = \frac{x^3}{3} \tan^{- 1} x - \frac{1}{6}\left( 1 + x^2 \right) + \frac{1}{6}\text{ log }\left| 1 + x^2 \right| + C\]
\[ = \frac{x^3}{3} \tan^{- 1} x - \frac{x^2}{6} + \frac{1}{6}\text{ log} \left| x^2 + 1 \right| + C'\text{  Where C' = C }- \frac{1}{6}\]